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MHTCET 10 May 2024 Shift 2 Paper

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Question : 120 of 150

Marks: +1, -0

If

\mathrm{w}=\frac{-1-\mathrm{i} \sqrt{3}}{2}

\mathrm{i}=\sqrt{-1}

, then the value of

\left|\begin{array}{ccc}1 & w & w^{2} \\ w & w^{2} & 1 \\ w^{2} & 1 & w\end{array}\right|

is

-1
0
1
3

Solution:

Concept:

Here,

\mathrm{w}

is a non-real cube root of unity.

So,

\mathrm{w}^{3}=1

and

1+\mathrm{w}+\mathrm{w}^{2}=0

.

Explanation:

The given determinant is

\begin{vmatrix} 1 & \mathrm{w} & \mathrm{w}^{2} \\ \mathrm{w} & \mathrm{w}^{2} & 1 \\ \mathrm{w}^{2} & 1 & \mathrm{w} \end{vmatrix}

.

Apply the row operation

R_{1} \rightarrow R_{1}+R_{2}+R_{3}

.

Each entry of the first row becomes

1+\mathrm{w}+\mathrm{w}^{2}=0

.

Therefore, the first row becomes entirely zero.

A determinant with a zero row has value

0

.

Answer:

The value of the determinant is

0

.

Hence, the correct option is B. 0.

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